3. The attempt at a solution
Mostly looking for some quick input if I'm getting this roughly right or not since there's no answer and I'm a bit unsure.
I first write ##f(\theta) = \sum_{-n}^n c_k e^{ik\theta}##. We're allowed to differentiate since ##f(\theta)## is in ##C^\infty##. Differentiating we have ##f'(\theta) = \sum_{-n}^n ikc_k e^{ik\theta}## and ##f''(\theta ) = \sum_{-n}^n -k^2c_k e^{ik\theta}##. every part of ##f(\theta )## is ##2\pi##-periodic so the only time we get a contribution to the integral is when the the exponential terms cancel each other exactly.
A) In this case when the index matches so we have##\langle Af, g\rangle = \int_{-\pi}^\pi \sum_{-n}^n \left( ika_kb_k \right) d\theta = 2\pi \sum_{-n}^n ika_k\overline{b_k}##.
and ##\langle f, Ag\rangle = \int_{-\pi}^\pi \left( \sum_{-n}^n -ika_k\overline{b_k} \right) d\theta = -2\pi \sum_{-n}^n ika_k\overline{b_k}## so not self-adjoint.
Similary we see that both B and C are self-adjoint.

The eigenvalues would be the functions ##e^{ikx}## for ##k = \pm 1, \pm 2,\dots , \pm n## and the eigenvalues for for ##A,B## and ##C## are ##ik, -k## and ##-k^2## respectively.

The case ##D## is slightly harder we have
##\langle Df,g \rangle = \langle f,g \rangle + \int_{-\pi}^\pi f(-\theta)\overline{g(\theta)}d\theta## and
##\langle f,Dg \rangle = \langle f,g \rangle + \int_{-\pi}^\pi f(\theta)\overline{g(-\theta)}d\theta##
So we need to prove that ##\int_{-\pi}^\pi f(-\theta)\overline{g(\theta)} = \int_{-\pi}^\pi f(\theta)\overline{g(-\theta)}d\theta##. Again matching the coefficients we have for the left side
##\int_{-\pi}^\pi \left( \sum_{-n}^n a_k\overline{b_{n-k}} \right) d\theta##
and the right side
##\int_{-\pi}^\pi \left( \sum_{-n}^n a_k\overline{b_{n-k}}\right) d\theta## which matches so self adjoint. The eigenvalues should be ##\frac{a_k+a_{n-k}}{a_k}## and the eigenfunctions the same as before. Did I understand this right?

The eigenvalues won't be of the form you gave because the ##a_k## coefficients are properties of the function ##f##, not the operator D.

Given the revised integral, can you work out what eigenvalues and eigenfunctions of D will be?

Thanks for taking the time replying! You're absolutely right about the index in that sum and thanks for clarifying that I can't have those constants from the function as an eigenvalue.

If I write out ##Df(\theta)## I have ##\sum_{-N}^N \left(c_k e^{ik\theta} + c_ke^{-ik\theta} \right) = \sum_{-N}^N (c_k + c_{-k})e^{ik\theta}##
I'm not sure I get all eigenfunctions here but I was thinking if I take eigenfunctions of the form ##f\in P_n## with the added constraint that ##c_k+c_{-k}=0## I have the eigenvalue zero?

I also possibly see other eigenvalues For example
if I choose the constrant that ##c_{-k} = ac_{k}## with ##a## being a complex number I have the eigenvalue ##(1+a)##. But not sure if I'm allowed too do this? The eigenvalue doesn't vary like in the last case(which was obviously wrong) but I do use properties from the function I guess.

That sounds right. What would be a neat basis for the eigenspace with eigenvalue zero?

Perhaps ##\{e^{ik\theta}-e^{-ik\theta} \}_0^n ## that is the basis ##\{ 1, e^{i\theta}-e^{-i\theta}, \dots , e^{in\theta}-e^{-in\theta} \}##.
The basis vector aint orthogonal nor normalized but if one wants we could get there with Gram-Schmidts method.

To see if this works, apply D to ##f(\theta)=e^{-ik\theta}+ae^{ik\theta}## and see what you get. Is it a complex scalar multiple of ##f(\theta)##?

What about other nonzero eigenvalues?

##D\left( e^{-ik\theta}+ae^{ik\theta} \right) = e^{-ik\theta} + ae^{ik\theta} + e^{ik\theta} + e^{-ik\theta} = (1+a) \left(e^{-ik\theta}+e^{ik\theta} \right)##. Right this doesn't work since I should have a ##a## in there as well.

So instead let's look at the case when ##c_{-k} = c_{k}## we have
##D(e^{ik\theta}+ e^{-ik\theta}) =e^{ik\theta}+ e^{-ik\theta} + e^{-ik\theta}+ e^{ik\theta} = 2(e^{ik\theta}+ e^{-ik\theta})##. So we have the eigenvalue two for the eigenfunctions as a linear combination of ##\{e^{ik\theta}+e^{-ik\theta} \}_0^n##. I don't think there's actually any more eigenvalues.